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Find the limit $$\lim_{x\to \frac{1}{\sqrt2}}\frac{\cos(\sin^{-1}x)-x}{1-\cot(\cos^{-1}x)}$$
First, I tried solving this sum using a substitute $\sin^{-1}x=t$, So when $x=\frac{1}{\sqrt{2}}$, $t=\frac{\pi}{4}$
$$\lim_{t\to \frac{\pi}{4}}\frac{\cos(t)-\sin(t)}{1-?}$$
But then I have trouble finding a value for $\cos^{-1}x$ in terms of t. Is my path correct? How should I proceed? Any hint would be highly appreciated... Thanks :)
P.S: I prefer solutions without using L'Hopital rule.
If $\sin^{-1}x=t\implies\cos^{-1}x=\frac\pi2-t$, therefore, $$\lim_{t\to\frac\pi4}\frac{\cos t-\sin t}{1-\cot(\frac\pi2-t)}=\lim_{t\to\frac\pi4}\frac{\cos t-\sin t}{1-\tan t}=\cos\frac\pi4$$
Hint:
Note that $\theta=\sin^{-1}x$ and $\alpha=\cos^{-1}x$.
According to image, can you find the value of $\cos(\sin^{-1}x)$ and $\cot(\cos^{-1}x)$?
We can find from the right triangle that: $$ \cos(\arcsin(x))=\sqrt{1-x^{2}},\tag{1} $$ and $$ \cot(\arccos(x))=\frac{x}{\sqrt{1-x^{2}}},\tag{2} $$ Therefore, $$ \lim_{x\to \frac{1}{\sqrt{2}}}\frac{\cos(\arcsin(x))-x}{1-\cot(\arccos(x))}=\lim_{x\to \frac{1}{\sqrt{2}}}\frac{\sqrt{1-x^{2}}-x}{1-\frac{x}{\sqrt{1-x^{2}}}}=\lim_{x\to \frac{1}{\sqrt{2}}}(\sqrt{1-x^{2}})=\frac{1}{\sqrt{2}}. $$
